LeetCode #1247 — MEDIUM

Minimum Swaps to Make Strings Equal

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

You are given two strings s1 and s2 of equal length consisting of letters "x" and "y" only. Your task is to make these two strings equal to each other. You can swap any two characters that belong to different strings, which means: swap s1[i] and s2[j].

Return the minimum number of swaps required to make s1 and s2 equal, or return -1 if it is impossible to do so.

Example 1:

Input: s1 = "xx", s2 = "yy"
Output: 1
Explanation: Swap s1[0] and s2[1], s1 = "yx", s2 = "yx".

Example 2:

Input: s1 = "xy", s2 = "yx"
Output: 2
Explanation: Swap s1[0] and s2[0], s1 = "yy", s2 = "xx".
Swap s1[0] and s2[1], s1 = "xy", s2 = "xy".
Note that you cannot swap s1[0] and s1[1] to make s1 equal to "yx", cause we can only swap chars in different strings.

Example 3:

Input: s1 = "xx", s2 = "xy"
Output: -1

Constraints:

1 <= s1.length, s2.length <= 1000
s1.length == s2.length
s1, s2 only contain 'x' or 'y'.

Step 01

Brute Force Baseline

Problem summary: You are given two strings s1 and s2 of equal length consisting of letters "x" and "y" only. Your task is to make these two strings equal to each other. You can swap any two characters that belong to different strings, which means: swap s1[i] and s2[j]. Return the minimum number of swaps required to make s1 and s2 equal, or return -1 if it is impossible to do so.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Greedy

Example 1

"xx"
"yy"

Example 2

"xy"
"yx"

Example 3

"xx"
"xy"

Core Insight

What unlocks the optimal approach

First, ignore all the already matched positions, they don't affect the answer at all. For the unmatched positions, there are three basic cases (already given in the examples):
("xx", "yy") => 1 swap, ("xy", "yx") => 2 swaps
So the strategy is, apply case 1 as much as possible, then apply case 2 if the last two unmatched are in this case, or fall into impossible if only one pair of unmatched left. This can be done via a simple math.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1247: Minimum Swaps to Make Strings Equal
class Solution {
    public int minimumSwap(String s1, String s2) {
        int xy = 0, yx = 0;
        for (int i = 0; i < s1.length(); ++i) {
            char a = s1.charAt(i), b = s2.charAt(i);
            if (a < b) {
                ++xy;
            }
            if (a > b) {
                ++yx;
            }
        }
        if ((xy + yx) % 2 == 1) {
            return -1;
        }
        return xy / 2 + yx / 2 + xy % 2 + yx % 2;
    }
}

// Accepted solution for LeetCode #1247: Minimum Swaps to Make Strings Equal
func minimumSwap(s1 string, s2 string) int {
	xy, yx := 0, 0
	for i := range s1 {
		if s1[i] < s2[i] {
			xy++
		}
		if s1[i] > s2[i] {
			yx++
		}
	}
	if (xy+yx)%2 == 1 {
		return -1
	}
	return xy/2 + yx/2 + xy%2 + yx%2
}

# Accepted solution for LeetCode #1247: Minimum Swaps to Make Strings Equal
class Solution:
    def minimumSwap(self, s1: str, s2: str) -> int:
        xy = yx = 0
        for a, b in zip(s1, s2):
            xy += a < b
            yx += a > b
        if (xy + yx) % 2:
            return -1
        return xy // 2 + yx // 2 + xy % 2 + yx % 2

// Accepted solution for LeetCode #1247: Minimum Swaps to Make Strings Equal
struct Solution;

impl Solution {
    fn minimum_swap(s1: String, s2: String) -> i32 {
        let n = s1.len();
        let s1: Vec<char> = s1.chars().collect();
        let s2: Vec<char> = s2.chars().collect();
        let mut x = 0;
        let mut y = 0;
        for i in 0..n {
            if s1[i] == 'x' && s2[i] == 'y' {
                x += 1;
            }
            if s1[i] == 'y' && s2[i] == 'x' {
                y += 1;
            }
        }
        if (x + y) % 2 != 0 {
            return -1;
        }
        x / 2 + y / 2 + x % 2 * 2
    }
}

#[test]
fn test() {
    let s1 = "xx".to_string();
    let s2 = "yy".to_string();
    let res = 1;
    assert_eq!(Solution::minimum_swap(s1, s2), res);
    let s1 = "xy".to_string();
    let s2 = "yx".to_string();
    let res = 2;
    assert_eq!(Solution::minimum_swap(s1, s2), res);
    let s1 = "xx".to_string();
    let s2 = "xy".to_string();
    let res = -1;
    assert_eq!(Solution::minimum_swap(s1, s2), res);
    let s1 = "xxyyxyxyxx".to_string();
    let s2 = "xyyxyxxxyx".to_string();
    let res = 4;
    assert_eq!(Solution::minimum_swap(s1, s2), res);
}

// Accepted solution for LeetCode #1247: Minimum Swaps to Make Strings Equal
function minimumSwap(s1: string, s2: string): number {
    let xy = 0,
        yx = 0;

    for (let i = 0; i < s1.length; ++i) {
        const a = s1[i],
            b = s2[i];
        xy += a < b ? 1 : 0;
        yx += a > b ? 1 : 0;
    }

    if ((xy + yx) % 2 !== 0) {
        return -1;
    }

    return Math.floor(xy / 2) + Math.floor(yx / 2) + (xy % 2) + (yx % 2);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.